Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{r^2 - 8r}{r^2 + 7r + 12} \times \dfrac{2r + 8}{r - 8} $
Solution: First factor the quadratic. $x = \dfrac{r^2 - 8r}{(r + 4)(r + 3)} \times \dfrac{2r + 8}{r - 8} $ Then factor out any other terms. $x = \dfrac{r(r - 8)}{(r + 4)(r + 3)} \times \dfrac{2(r + 4)}{r - 8} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ r(r - 8) \times 2(r + 4) } { (r + 4)(r + 3) \times (r - 8) } $ $x = \dfrac{ 2r(r - 8)(r + 4)}{ (r + 4)(r + 3)(r - 8)} $ Notice that $(r - 8)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 2r(r - 8)\cancel{(r + 4)}}{ \cancel{(r + 4)}(r + 3)(r - 8)} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $x = \dfrac{ 2r\cancel{(r - 8)}\cancel{(r + 4)}}{ \cancel{(r + 4)}(r + 3)\cancel{(r - 8)}} $ We are dividing by $r - 8$ , so $r - 8 \neq 0$ Therefore, $r \neq 8$ $x = \dfrac{2r}{r + 3} ; \space r \neq -4 ; \space r \neq 8 $